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3 years ago

Which is the product of 14 and 3/7

Mathematics

2 answers:

nasty-shy [4]3 years ago

4 0

6

3/7 x 14/1 =6
This could be done with a Calculator

zysi [14]3 years ago

4 0

Answer: 42/7 = 6

Multiply Numerator by Numerator

14*3=42

Multiply Denominator by Denominator

1*7=7

Get your answer

42/7

Divide

42/7=6

Final answers

42/7 or 6

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Luis, Marlon, and Victoria are counting money. They each have the same amount in dollar bills, quarters, dimes, nickels, and pen

yaroslaw [1]

I believe the correct answer from the choices listed above is the third option. The exercise best illustrates Medium of exchange and portability function and characteristic of money. 

Hope this answers the question. Have a nice day.

6 0

3 years ago

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The price of a four-star meal, only $5050 per couple decades ago, has since increased 400400%. The current price of a four-

Vikentia [17]

Answer:

The correct answer is C.

Step-by-step explanation:

The price of a four-star meal, only $50 per couple decades ago, has since increased 400%.

This means that if the $50 per couple is 100%(decimal 1), the current price is 100%+400% = 500%(decimal 5). So we can solve a rule of three to find the current price of a four-star meal.

1 - 50

5 - x

$x = 50*5$

$x = 250$ .

The current price of a four star meal is $250 per couple.

The correct answer is C.

3 0

3 years ago

Find all of the eigenvalues λ of the matrix A. (Hint: Use the method of Example 4.5 of finding the solutions to the equation 0 =

Svetradugi [14.3K]

Answer:

$\lambda=8,\ \lambda=-5$

Step-by-step explanation:

Eigenvalues of a Matrix

Given a matrix A, the eigenvalues of A, called $\lambda$ are scalars who comply with the relation:

$det(A-\lambda I)=0$

Where I is the identity matrix

$I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

The matrix is given as

$A=\left[\begin{array}{cc}3&5\\8&0\end{array}\right]$

Set up the equation to solve

$det\left(\left[\begin{array}{cc}3&5\\8&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda \end{array}\right]\right)=0$

Expanding the determinant

$det\left(\left[\begin{array}{cc}3-\lambda&5\\8&-\lambda\end{array}\right]\right)=0$

$(3-\lambda)(-\lambda)-40=0$

Operating Rearranging

$\lambda^2-3\lambda-40=0$

Factoring

$(\lambda-8)(\lambda+5)=0$

Solving, we have the eigenvalues

$\boxed{\lambda=8,\ \lambda=-5}$

8 0

3 years ago

Please help me with my math

Natasha_Volkova [10]

Answer:

The answer is C

Step-by-step explanation:

the y-intercept is going through 0 therefore there would be no "b" in the y=mx+b formula

3 0

2 years ago

Help me guys thx so much

navik [9.2K]

Answer:

Option D, $10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

Step-by-step explanation:

Step 1: Multiply

$(\sqrt{10x^4} -x\sqrt{5x^2} )*(2\sqrt{15x^4} + \sqrt{3x^3})\\ (\sqrt{10 * x^2 * x^2} -x\sqrt{5 * x^2} ) * (2\sqrt{15 * x^2 * x^2} +\sqrt{3 * x^2 * x})\\(x^2\sqrt{10} -x^2\sqrt{5} )*(2x^2\sqrt{15} +x\sqrt{3x}) \\\\$

$(x^2\sqrt{10}*2x^2\sqrt{15} )+(x^2\sqrt{10}*x\sqrt{3x} ) + (-x^2\sqrt{5} *2x^2\sqrt{15}) + (-x^2\sqrt{5} *x\sqrt{3x}$

$(2x^4\sqrt{150} ) + (x^3\sqrt{30x}) + (-2x^4\sqrt{75}) + (-x^3\sqrt{15x} )$

$2x^4\sqrt{5^2*6} + x^3\sqrt{30x} -2x^4\sqrt{5^2*3} -x^3\sqrt{15x}$

$10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

Answer: Option D, $10x^4\sqrt{6} +x^3\sqrt{30x} -10x^4\sqrt{3} -x^3\sqrt{15x}$

6 0

3 years ago

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